Q70 / 180 · KCET 2010 · Mathematics

If $x+i y=(-1+i \sqrt{3})^{2010}$, then $x$ is

  • A
    $-2^{2010}$
  • Correct answer
    B
    $2^{2010}$
  • C
    1
  • D
    $-1$

Solution

$x+i y=(-1+i \sqrt{3})^{2010}$ $\Rightarrow \quad x+i y=(2)^{2010}\left(\frac{-1+i \sqrt{3}}{2}\right)^{2010}$ $\Rightarrow \quad(x+i y)=(2)^{2010} \omega^{2010} \quad\left(\because \omega=\frac{-1+i \sqrt{3}}{2}\right)$ $\Rightarrow \quad(x+i y)=(2)^{2010}\left(\omega^{3}\right)^{670} \quad\left(\because \omega^{3}=1\right)$ $\Rightarrow \quad(x+i y)=(2)^{2010}(1)^{670}=2^{2010}+i \cdot 0$ On comparing real part $\Rightarrow \quad \quad x=2010$

Hence, the correct answer is option: 'B'.

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